I am looking at the datasheet for the 74HC4511 (available here), and I am confused by the usage of two of the pins: Firstly, the "Latch Enable" pin (\$\overline{LE}\$), and secondly the "Ripple Blanking" pin (\$\overline{BI}\$).
Looking at the logic table for the IC gives states that if the Latch Enable, is high, and neither of the other two "special pins", \$\overline{BI}\$ and \$\overline{LT}\$ are active, then the output is dependent upon the BCD code applied during the transition of \$\overline{LE}\$ from low to high. Does this mean that it stores the BCD state and continues to output the corresponding 7-segment output (for instance if \$0001\$ was applied whilst \$\overline{LE}\$ was transitioning, and then was subsequently changed to \$0010\$, the 7-segment display would continue to display \$1\$ rather than \$2\$?).
I am more confused by the purpose of the Ripple Blanking input, it appears that this simply blanks the 7-segment display, but I do not know what the purpose of this would be, when you can simply set the BCD code to \$1111\$?)
Thanks in advance, and sorry if this is obvious!
Best Answer
If the 4-bit data input to one 4511 is being shared with another 4511, LE allows you to set-up the 4-bit data for one device (then store it on internal registers). Then you can set up the appropriate 4-bit data (using the same data lines) for another 4511 and have a separate LE pin to store this different data to the 2nd device.
If you didn't have this feature you would have to use 4 lines dedicated to each driver and that is inefficient.
The blanking input disables all segments.